I am calculating two different $p$ values for the McNemar test and what I believe is the corresponding exact test, even for combined counts larger than 25.

Take, for example the contingency table

Posttest
Pretest Outcome A Outcome B
Outcome A xx 50
Outcome B 65 xx

The chi-square statistic is
$$\chi^2=\frac{(50-65)^2}{50+65}=1.96,$$
with
$$p=0.1619. \tag{R}$$

According to Sheskin (2011, Test 20, VI.3, pg 844), the exact test for these situations is essentially a binomial sign test (for a single sample) with parameter $\pi=0.5$ and the two counts equal to the two the cells of interest in the contingency table. My understanding of this is that I'm calculating the probability that, out of $50+65=115$ trials with probability of success 0.5, at least 65 of them are successes. My attempt to calculate this for the particular situation above is:
$$
p
=
\sum_{i=65}^{115}
\begin{pmatrix} 115 \\ i \end{pmatrix}
0.5^{i}\ 0.5^{115-i}
=
0.0957
\tag{Mathematica}
$$

Why do these two $p$-values differ so much? Am I making a simple mistake?

2 Answers
2

McNemar's test uses discrete data (counts of discordant pairs) to produce a $\chi^{2}$ test statistic with one degree of freedom. However, the $\chi^{2}$ distribution is not discrete, but continuous. Hence, the need for a continuity correction.